How to get the second line of this equation from the first one?

[zb23]

TL;DR Summary

biot savart law,curl of vector potential

 

[kuruman]

It's a double integral. Anything that does not depend on double-prime is constant and can be pulled out of the integral over double primed variables.

 

[zb23]

Can you be a little more specific considering this case?

 

[kuruman]

There is nothing specific about this. Just put everything that depends on double prime under the integral sign that has ##dV''## and everything that depends only on prime under the integral sign that has ##dV'##. You do the double prime integral first and get a function of ##r'## and ##r##. Next you do the integral over primed variables and you get a function of ##r## only which will be an expression for ##Z(r)##.

It's like $$\int \int f(x,y)~g(x)~dx~dy=\int g(y)dy\int f(x,y)dx$$except with primes and double primes instead of ##x## and ##y##.

 

[zb23]

Ok.Thank you.

 

[vanhees71]

There is nothing specific about this. Just put everything that depends on double prime under the integral sign that has ##dV''## and everything that depends only on prime under the integral sign that has ##dV'##. You do the double prime integral first and get a function of ##r'## and ##r##. Next you do the integral over primed variables and you get a function of ##r## only which will be an expression for ##Z(r)##.

It's like $$\int \int f(x,y)~g(x)~dx~dy=\int g(y)dy\int f(x,y)dx$$except with primes and double primes instead of ##x## and ##y##.

There's something wrong. Shouldn't it be
$$\int \mathrm{d} x \int \mathrm{d} y f(x,y)g(y)=\int \mathrm{d} y \int \mathrm{d} x f(x,y)g(y) = \int \mathrm{d} y g(y) \int \mathrm{d} x f(x,y)?$$
Examples like this let me prefer to write the differential of the integral in front, i.e., to have the integral sign including the differential as an operator acting to the right. Usually this makes reading the integrals and manipulating them easier than the somehow more common notation in the math literature, which puts the differential at the very end of the expression, i.e., after the integrand.